The formula for the sum of a geometric sequence provides a great opportunity to illustrate different styles of proof.

The formula under discussion:

a + ax + ax^{2} + ax^{3} + ... + ax^{n} = a*(1-x^{n+1})/(1-x)

The most obvious way to prove a formula of this sort (one with an n in it) is by mathematical induction:

**Step 1** (Show it's true for n=0):
a = a*(1-x)/(1-x)
Yup... That's true.
**Step 2** (Show that if it's true for n, it's also true for n+1):
a + ax + ... + ax^{n} + ax^{n+1} =
(by assuming true for n) a*(1-x^{n+1}) / (1-x) + ax^{n+1}
(multiply by (1-x)/(1-x)) = a*(1-x^{n+1}) / (1-x) + ax^{n+1}(1-x)/(1-x)
(add fractions) = (a-ax^{n+1}+ax^{n+1}-ax^{n+2}) / (1-x)
= (a-ax^{n+2}) / (1-x)
And that's what we wanted to prove.

Although this is often considered to be a perfect example of when to use mathematical induction, it's not as though it's particularly revealing of any deeper truths. And it's easy to miss a minus sign in the algebra involved, so if you forget the formula and need to derive it again, not only do you have to guess the right one, but you also need to get all the minus signs in the right place.

So, in this case, a direct proof is more illustrative:

We'll call the sum we're looking for S, so
Let S = a + ax + ax^{2} + ... + ax^{n}
Then xS = ax + ax^{2} + ... + ax^{n} + ax^{n+1}
So, if we subtract xS from S, we get:
S - xS = a - ax^{n+1}
Factor out the a on the right side, and then divide by (1-x):
S = a * (1-x^{n+1}) / (1-x)

This proof even has a neat little mnemonic to help you remember the important step: "subtract off the excess (xS)". Okay, so it's not really that funny, but most puns aren't. At any rate, the math here is easier.

As an added bonus, it's really easy to get the sum of an infinite sequence from this formula. Just take the limit as n goes to infinity. There's only the one bit with an n in it (the x^{n+1}). If the absolute value of x is less than 1, this goes to 0. Otherwise, it goes to infinity. We're interested in cases that don't go to infinity, so we're left with the formula

S = a * (1-0) / (1-x) = a / (1-x)

iff
-1 < x < 1